# List all points where the max. and min. value of f(x,y) = 2x^3 + 2x^2y + \frac{4}{3}y^3 with the...

## Question:

List all points where the max. and min. value of

{eq}f(x,y) = 2x^3 + 2x^2y + \frac{4}{3}y^3 {/eq}

with the restriction {eq}x^2 + y^2 = 40 {/eq}, might occur.

## Equality Constrained Optimization using Lagrange Multipliers:

To find the critical points of the function {eq}f(x,y) {/eq} subject to the constraint {eq}g(x,y) = k {/eq}, we use the method of Lagrange multipliers. In this approach, the critical points are the solutions to the system of equations consisting of {eq}\nabla f(x,y) = \lambda \nabla g(x,y) {/eq} and {eq}g(x,y) = k {/eq} where {eq}\lambda {/eq} is the Lagrange multiplier. To classify the identified critical points, we evaluate each critical point using the objective function. The largest function value is the maximum and the smallest function value is the minimum.

## Answer and Explanation:

We must first generate the system of equations that incorporates the Lagrange multiplier and will yield the critical points.

$$\begin{align*} \nabla (2x^3 + 2x^2y + \frac{4}{3}y^3) &= \lambda \nabla (x^2 + y^2) \\ x^2 + y^2 &= 40 \end{align*} $$

To find the critical points, we have to find the gradient of the first equation of the system.

$$\begin{align*} 6x^2 + 4xy &= 2\lambda x \\ 2x^2 + 4y^2 &= 2\lambda y \\ x^2 + y^2 &= 40 \end{align*} $$

Now that we have the fully expanded system, we solve to find the critical points. We begin by finding the first two equations for {eq}\lambda {/eq} in terms of {eq}x {/eq} and {eq}y {/eq}.

$$\begin{align*} 6x^2 + 4xy &= 2\lambda x \\ \lambda &= \frac{6x^2 + 4xy}{2x} \\ 2x^2 + 4y^2 &= 2\lambda y \\ \lambda &= \frac{2x^2 + 4y^2}{2y} \\ \end{align*} $$

Then we set {eq}\lambda {/eq} equal to itself and find a relationship between {eq}x {/eq} and {eq}y {/eq}.

$$\begin{align*} \lambda &= \lambda \\ \frac{2x^2 + 4xy}{2x} &= \frac{2x^2 + 4y^2}{2y} \\ 4x^2y + 8xy^2 &= 4x^3 + 8xy^2 \\ 4x^2y &= 4x^3 \\ 0 &= 4x^3 - 4x^2y \\ 0 &= 4x^2(x - y) \\ x &= 0 \\ x &= y \end{align*} $$

Using the final equation of the system, we find the numerical solutions to the system.

$$\begin{align*} x^2 + y^2 &= 40 \\ \text{Let } x &= 0 \\ y^2 &= 40 \\ y &= \pm 2\sqrt{10} \\ \text{Let } x &= y \\ x^2 + x^2 &= 40 \\ 2x^2 &= 40 \\ x^2 &= 20 \\ x &= \pm 2\sqrt5 \end{align*} $$

The critical points of the function are {eq}(0, \pm 2\sqrt{10} {/eq} and {eq}(\pm 2\sqrt5, \pm 2\sqrt5) {/eq}.

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from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9